The figure shows a simple RC circuit consisting of a 10.0-MUF capacitor in serie

Question

The figure shows a simple RC circuit consisting of a 10.0-MUF capacitor in series with a resistor Initially, the switch is open as suggested in the figure The capacitor has been charged so that the potential difference between its plates is 100.0 V. At t =0 s, the switch is closed. The capacitor discharges exponentially so that 2.0 s after the switch is closed. the potential difference between the capacitor plates is 37 V. In other words, in 2.0 s the potential difference between the capacitor plates is reduced to 37 % of its original value Calculate the electric potential energy stored in the capacitor before the switch is closed

Explanation / Answer

before the switch is closed:
voltage across capacitor = 100V
capacitance = 10F

energy stored in capacitor : W = 1/2 (CV^2)

= 1/2 (10F)(100)^2 = 0.01 joules

i hope it helps you dear :))

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